The $\ce{N-N}$ bond length in $\ce{N2O4}$ is $\pu{1.64 Å}$ which is much longer than in the case of Hydrazine $\ce{N2H4}$ $\pu{(1.47 Å)}$. This means that the bond is quite weak. Another unusual case I see that $\ce{N-N}$ bond length in $\ce{N2O3}$ ( the blue asymmetric form ) is $\pu{1.86 Å}$ !!

That is very weak for a Nitrogen single bond. Is there a satisfactory explanation for the same? And amazingly molecules with such long bonds tend to have a nonplanar structure due to bond rotation but $\ce{N2O4}$ is planar.

I found a similar question on N2O4 but the answer to the planarity question wasn't so satisfactory. Also, I wanted the extend upon the weak bond question.


3D space filling model of N2O4

3D space filling model of N2H4

Images are from Wikipedia.

The first obvious thing that you can see is that N2O4 is planar but N2H4 is not. And the primary reason starts here.

N2O4 can be viewed as two nitrogen dioxide molecules joined together. If you draw a Lewis structure of NO2, you will recognize two things. Firstly, it has one unpaired electron. So imagine when two NO2 molecules come near together, they will share their two unpaired electrons (like how radicals react that you might be familiar with in organic chemistry) to form a single sigma bond. This is the very bond you are asking about.

And secondly, it is sp2 hybridized at the nitrogen atom. Thus, when you join the two NO2 molecules together, taking the relative orientation of the sp2 orbitals and how they should overlap to form a sigma bond in mind, you can see why N2O4 is planar. Because the N=O bonds (I write the "=" symbol for double bond for simplicity, it should be about 1.5 order due to resonance) heavily repulse each other when they are in the planar position, it weakens the N-N bond. Indeed this bond is so weak that it is homolytically dissociated at near below room temperature or above, creating basis for the popular equilibria between NO2 and N2O4.

The situation is even worse for N2O3. N2O3 can be viewed as formed from a NO (itself has an unpaired electron and a sp2 nitrogen) and a NO2 molecule, with the exact mechanism as above. And guess what? N2O3 has a lone electron pair on the nitrogen (in place of the missing oxygen compared to N2O4). Lone pairs repulse much, much more heavily than the bonding pairs do, and this further weakens the N2O3 N-N bond. This is illustrated by the easily disproportionation of N2O3 into NO and NO2.

Then why N2H4's bond is stronger? Because the nitrogen atoms are sp3 hybridized in this case (note that oxygen is divalent but hydrogen is only monovalent), thus they form a tetrahedral electronic structure. Thus they can comfortably adopt a skew conformation to minimize the repulsion. Either N2O3 or N2O4 could achieve this because the sp2 hybridization confines them to planar geometry.

Using Pauling's formula with the 1.14 A triple bond in N2 as a reference:

D(n) = D(m) - 0.6log(n/m)

With D(n, m) is the length of the bond with order n, m respectively.

The N-N bond order in N2O4, N2O3 and N2H4 can be estimated to be 0.44, 0.19 and 0.85 respectively.

  • 1
    $\begingroup$ "you can see why N2O4 is planar - Sorry, but no I can't, especially if the N-N bond is purely a sigma bond. Further the iso-electronic oxalate ion can be found in both planar and non-planar forms (en.wikipedia.org/wiki/Oxalate). I agree that iff N2O4 and similar are planar then repulsion between the oxygens is a good hand waving argument to explain the long bond, the problem is that very argument suggests to me that it should not be planar in the first place. So could you expand on why this explains the planarity of the species in the first place? $\endgroup$
    – Ian Bush
    May 26 '21 at 19:46
  • $\begingroup$ I'm seeing $D_{\rm 2h}$ symmetry for the $s$-$\ce{N2O4}$ dimer here - as to the why, that's another kettle of fish. $\endgroup$
    – Todd Minehardt
    May 26 '21 at 20:57
  • $\begingroup$ @Ian Bush this paper might help: onlinelibrary.wiley.com/doi/abs/10.1002/jcc.540020106 $\endgroup$ May 27 '21 at 5:02
  • $\begingroup$ @ĐỨcLêHồng Unfortunately where I work doesn't let me through the paywall. But the abstract seems to confirm my own rationalisation for why N2O4 is planar: "Experiment says it is planar. Good level theory says it is planar. Accept it - Chemistry is difficult, hand waving doesn't solve everything" $\endgroup$
    – Ian Bush
    May 27 '21 at 6:46
  • $\begingroup$ @ĐỨcLêHồng Thanks for the opinion, that bond order thing makes it quite clear. $\endgroup$
    – Swadhin
    May 27 '21 at 9:43

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